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Đặt A = 1/2.6 + 1/6.10 + 1/10.14 + ..... + 1/102.106
=> 4A = 4/2.6 + 4/6.10 + 4/10.14 + ..... + 4/102.106
=> 4A = 1/2 - 1/6 + 1/6 - 1/10 + 1/10 - 1/14 + ... + 1/102 - 1/106
=> 4A = 1/2 - 1/106
=> 4A = 26/53
=> A = 13/106
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\(\frac{1}{2.6}+\frac{1}{6.10}+...+\frac{1}{102.106}\)
\(=\frac{1}{4}.\left(\frac{4}{2.6}+\frac{4}{6.10}+...+\frac{4}{102.106}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+...+\frac{1}{102}-\frac{1}{106}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{106}\right)\)
\(=\frac{1}{4}.\frac{26}{53}\)
\(=\frac{13}{106}\)
12.T=2.6.12+6.10.12+10.14.12+...+102.106.12=
=2.6.(10+2)+6.10.(14-2)+10.14.(18-6)+...+102.106.(110-98)=
=2.2.6+2.6.10-2.6.10+6.10.14-6.10.14+10.14.18-...-98.102.106+102.106.110=
=2.2.6+102.106.110
\(\Rightarrow T=\dfrac{2.2.6+102.106.110}{12}=99112\)
a: \(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{201}-\dfrac{1}{203}=\dfrac{202}{203}\)
b: \(=-4\left(\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+...+\dfrac{1}{2015\cdot2018}\right)\)
\(=-\dfrac{4}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{2015\cdot2018}\right)\)
\(=\dfrac{-4}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{2015}-\dfrac{1}{2018}\right)\)
\(=\dfrac{-4}{3}\cdot\dfrac{504}{1009}=-\dfrac{672}{1009}\)
a) 2/2.5 + 2/5.8 + 2/8.11 + ... + 2/x(x+3) = 7/23
3/2.5 + 3/5.8 + 3/8.11 + ... + 3/x(x+3) = 21/46
1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/x - 1/x+1 = 21/46
1/2 - 1/x+1 = 21/46
=> 1/x+1 = 1/23
=> x + 1 = 23
=> x = 22
Vậy x = 22.
b) 3/4 . x - 1/5 = 7/4 . x + 11/5
3/4 . x - 7/4 . x = 1/5 + 11/5
x (3/4 - 7/4) = 12/5
-x = 12/5
x = -12/5
Vậy x = -12/5.
Đề hình như bị sai ban ơi sửa lại
\(A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{92.95}\)
\(A=3\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}\right)\)
\(A=3.\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}\right)\)
\(A=\dfrac{1}{2}-\dfrac{1}{95}\)
\(A=\dfrac{93}{190}\)
\(B=\dfrac{2}{2.5}+\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{92.95}\)
\(3B=2\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}\right)\)
\(3B=2.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}\right)\)
\(3B=2\left(\dfrac{1}{2}-\dfrac{1}{95}\right)\)
\(3B=2.\dfrac{93}{190}\)
\(3B=\dfrac{93}{95}\)
\(\Rightarrow B=\dfrac{31}{95}\)
\(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2015.2018}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2015}-\frac{1}{2018}\)
\(=\frac{1}{2}-\frac{1}{2018}=\frac{504}{1009}\)
Vậy \(A=\frac{504}{1009}.\)
\(B=\frac{4}{2.6}+\frac{4}{6.10}+\frac{4}{10.14}+...+\frac{4}{102.106}\)
\(=\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{102}-\frac{1}{106}\)
\(=\frac{1}{2}-\frac{1}{106}=\frac{26}{53}\)
Vậy \(B=\frac{26}{53}.\)
Bài làm:
a) \(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2015.2018}\)
\(A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2015}-\frac{1}{2018}\)
\(A=\frac{1}{2}-\frac{1}{2018}\)
\(A=\frac{504}{1009}\)
b) \(B=\frac{4}{2.6}+\frac{4}{6.10}+\frac{4}{10.14}+...+\frac{4}{102.106}\)
\(B=\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{102}-\frac{1}{106}\)
\(B=\frac{1}{2}-\frac{1}{106}\)
\(B=\frac{26}{53}\)